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# LineContainer

Authors: Benjamin Qi, Andi Qu

### Prerequisites

Convex Containers

## Half-Plane Intersection

Resources
CF
Petr

Expected linear time!

### This section is not complete.

Any help would be appreciated! Just submit a Pull Request on Github.
Example problem + implementation?
StatusSourceProblem NameDifficultyTags
KattisNormal
JOIHard
Balkan OIVery Hard
Show TagsBinary Search, Geometry

## LineContainer (aka $\mathcal{O}(N \log N)$ CHT)

StatusSourceProblem NameDifficultyTags
YSNormal
Resources
KACTL

source of code that I (Ben) use

cp-algo

related topic (but not the same)

### Example Problem

Focus Problem – read through this problem before continuing!

#### Analysis

Instead of focusing on the pillars that should be destroyed, let's instead focus on the pillars that remain.

The total cost consists of the cost due to height differences plus the cost of destroying unused pillars. The latter cost is equal to the cost to destroy all pillars minus the cost to destroy the remaining pillars.

Since the cost to destroy all pillars is constant, we can thus turn the problem into one about building pillars instead of destroying them!

From this, we get a basic DP recurrence. Let $dp[i]$ be the minimum cost to build the bridge so that the last build pillar is pillar $i$.

$dp[1] = -w_1$ and the following recurrence holds:

\begin{aligned} dp[i] &= \min_{j < i}(dp[j] + (h_j - h_i)^2 - w_i)\\ &= \min_{j < i}(dp[j] + h_j^2 - 2h_ih_j) + h_i^2 - w_i \end{aligned}

Notice how

$dp[j] + h_j^2 - 2h_ih_j$

effectively describes a linear function $y = mx + c$, where $m = -2h_j$, $x = h_i$, and $c = dp[j] + h_j^2$

This means that we can use CHT to compute $dp[i]$ efficiently!

However, since $m$ is not monotonic, we can't use linear CHT using a deque, so we must settle with $\mathcal{O}(N \log N)$.

Code

## Problems

StatusSourceProblem NameDifficultyTags
YSNormal
POINormal
Show TagsConvex, DP
CEOIHard
Show TagsConvex, DP
FHCHard
Show TagsConvex, DP
ACHard
TLXHard
Old GoldHard

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